Note: This is a lemma for: Spectral Measures: Riemann-Lebesgue
Given a positive measure: $$\lambda:\mathcal{A}\to[0,\infty]$$
Consider a complex measure: $$\mu:\mathcal{A}\to\mathbb{C}$$
How to prove the equivalence? $$\mu\ll\lambda\iff|\mu|\ll\lambda$$ (This can be quite helpful in some situations!)
Hint: There are actually a few proofs; here's one.
Positive measures preserve null sets: $$\lambda(N)=0:\quad\lambda(E)=0\quad(E\subseteq N)\implies\mu(E)=0\quad(E\subseteq N)$$ So the Radon-Nikodym must vanish here: $$E\subseteq N\quad0=\mu(E)=\int_E u\mathrm{d}|\mu|\implies u1_N=0\mod{|\mu|}$$ That in turn gives absolute continuity: $$|\mu|(N)=\int_N|u|\mathrm{d}|\mu|=0$$